Answer:
To find \( \frac{dy}{dx} \) and \( \frac{d^2y}{dx^2} \) for the given functions \( x = e^t \) and \( y = te^{-t} \), we'll use the chain rule.
First, let's find \( \frac{dy}{dt} \):
\( y = t e^{-t} \)
Using the product rule:
\( \frac{dy}{dt} = t \frac{d}{dt}(e^{-t}) + e^{-t} \frac{d}{dt}(t) \)
Now, we can find \( \frac{d}{dt}(e^{-t}) \) and \( \frac{d}{dt}(t) \):
\( \frac{d}{dt}(e^{-t}) = -e^{-t} \) (Derivative of \( e^{-t} \) with respect to \( t \))
\( \frac{d}{dt}(t) = 1 \) (Derivative of \( t \) with respect to \( t \))
Now, plug these values back into the expression for \( \frac{dy}{dt} \):
\( \frac{dy}{dt} = t(-e^{-t}) + e^{-t}(1) \)
\( \frac{dy}{dt} = -te^{-t} + e^{-t} \)
Now, let's find \( \frac{dx}{dt} \):
\( x = e^t \)
\( \frac{dx}{dt} = \frac{d}{dt}(e^t) \)
\( \frac{dx}{dt} = e^t \)
Now, we can use the chain rule to find \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} \)
\( \frac{dy}{dx} = (-te^{-t} + e^{-t}) \cdot (e^t) \)
\( \frac{dy}{dx} = (-t + 1) \)
Now, let's find \( \frac{d^2y}{dt^2} \) by differentiating \( \frac{dy}{dt} \) with respect to \( t \):
\( \frac{d}{dt}\left(-te^{-t} + e^{-t}\right) = -\frac{d}{dt}(te^{-t}) + \frac{d}{dt}(e^{-t}) \)
Using the product rule for the first term:
\( -\frac{d}{dt}(te^{-t}) = -t\left(-e^{-t}\right) + \frac{d}{dt}(e^{-t}) \)
Now, plug in the derivatives we found earlier:
\( -\frac{d}{dt}(te^{-t}) = te^{-t} + e^{-t} \)
So,
\( \frac{d^2y}{dt^2} = te^{-t} + e^{-t} \)
Now, to find \( \frac{d^2y}{dx^2} \), we can use the chain rule again:
\( \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) \)
\( \frac{d^2y}{dx^2} = \frac{d}{dx}(-t + 1) \)
\( \frac{d^2y}{dx^2} = -\frac{d}{dx}(t) + \frac{d}{dx}(1) \)
Since \( t \) is a constant with respect to \( x \), its derivative with respect to \( x \) is zero:
\( \frac{d}{dx}(t) = 0 \)
So,
\( \frac{d^2y}{dx^2} = 0 \)
Therefore, \( \frac{d^2y}{dx^2} = 0 \) for the given functions \( x = e^t \) and \( y = te^{-t} \).
Explanation: